Calculate the Electric Potential

In summary, the conversation involves a problem where the electric potential at point B, located on the perpendicular bisector of a rod a distance b above the x-axis, needs to be calculated. The rod is described as having a varying charge density [lamb] = [alpha]x, where [alpha] is a constant. The correct answer for the potential is V = -(k[alpha]L/2)ln{[squ] [(L^2/4) + b^2] + L/2)/[squ] [(L^2/4) + b^2] - L/2}, but the person is not getting the same answer. They have set up the integral correctly, but made a mistake in
  • #1
discoverer02
138
1
What am I doing wrong here?

Here's the problem:
For the arrangement descripbed in the previous problem (see attachment), calculate the electric potential at point B that lies on the perpendicular bisector of the rod a distance b above the x axis.

[lamb] = [alpha] x where [alpha] is a constant.

The correct answer is
V = -(k[alpha]L/2)ln(([squ] [(L^2/4) + b^2)] - L/2)/[squ] [(L^2/4) + b^2)] - L/2))

I'm not getting the same answer.

So far I've got the following: [the] = [<]'a' on the diagram.

V = kq/r
x' = btan[the]
dx' = bsec^2[the]d[the]
x = L/2 + x'
r = bsec[the]

dq = [lamb]d(L/2 + x') = [lamb]dx'
dq = [alpha](L/2 + x')dx' = [alpha](L/2 + btan[the])bsec^2[the]d[the]

so dV = k[alpha](L/2 + btan[the])bsec^2[the]d[the]/bsec[the]

Taking the integral of both sides from -[the] to +[the] doesn't yield the correct result.

I'd appreciate it if someone could point out where I went wrong. I have a feeling the problem's in [lamb] = dq/dL = dq(L/2 + x').

Thanks.
 
Physics news on Phys.org
  • #2
I forgot the attachment. Here it is.
 

Attachments

  • diagram.bmp
    23.7 KB · Views: 711
  • #3
How is that rod charged?

You say
dq = &lambda; dx' with
&lambda; = &alpha; x.

This would mean that the rod is NOT uniformly charged. Are you sure this is what the problem says?

Normally, problems like this have a uniformly charged rod, meaning
sth. like dq = &lambda; dx', where &lambda; is a constant.
 
  • #4
The rod is not uniformly charged. [lamb] varies with x.
 
  • #5
Small typo in the answer. Should be:

-(k&alpha;L/2)ln{( sqrt[(L2/4) + b2)] + L/2)/ (sqrt[(L2/4) + b2] - L/2)}
(the answer you posted equals 0)

Don't set it up in terms of the angle.
You want V = -k&int;dq/r
dq = &Lambda;dx and &Lambda; = &alpha;x
therefore:
dq = &alpha;xdx
and
r = sqrt(b2 + (x - L/2)2)

so your integral is

V = -k&alpha;&int; (xdx/(sqrt(b2 + (x - L/2)2)) from x=0 to x=L

It actually does work out. Happy integrating. :smile:
 
Last edited:
  • #6
Thanks again gnome.
 
  • #7
That integral is a royal pain!

I'm curious. What was wrong with the integral I was originally working with other than the fact that it might be easier integrating with respect to 'x' rather than the angle?
 
  • #8
Actually, nothing. Shoulda done it your way.

dV = k&alpha;(L/2 + btan&theta;)bsec^2&theta;d&theta;/bsec&theta;
dV = k&alpha;(L/2 + btan&theta;)sec&theta;d&theta;
V = k&alpha;L/2 &int;sec&theta;d&theta; + k&alpha;b &int;tan&theta;sec&theta;d&theta;
V = k&alpha;L/2 {ln|sec&theta; + tan&theta;|}{from -&theta; to &theta;} + k&alpha;b{sec&theta;}{from -&theta; to &theta;}
The way you have defined theta, up at the top there, sec(-&theta;) = sec(&theta;) = (sqrt[L2 + 4b2])/(2b) so the second term of the integral drops out completely.

tan&theta; = L/(2b)
tan(-&theta;) = -L/(2b)

so the integral, evaluated from -&theta; to &theta;, becomes:
(k&alpha;L/2)(ln{(sqrt[L2 + 4b2])/(2b) + L/(2b)} - ln{(sqrt[L2 + 4b2])/(2b) - L/(2b)})

Play around with that a little & you get exactly the same answer.
 
  • #9
Thanks.

It's a relief to know I set the integral up correctly. A stupid mistake on my part evaluating the integral turned into something it's not, so I thought I was going about it the wrong way.
 

FAQ: Calculate the Electric Potential

What is electric potential?

Electric potential is a measure of the electric potential energy per unit of charge at a specific point in an electric field. It represents the amount of work needed to move a unit of charge from an infinitely far away point to the specific point in the electric field.

How is electric potential calculated?

Electric potential is calculated by dividing the electric potential energy by the amount of charge present at a specific point in an electric field. It can also be calculated by multiplying the electric field strength by the distance between the two points.

What are the units of electric potential?

The units of electric potential are joules per coulomb (J/C) in the SI system. In some cases, it is also expressed in volts (V), which is equivalent to joules per coulomb.

How is electric potential different from electric potential energy?

Electric potential is a measure of the potential energy per unit of charge at a specific point in an electric field, while electric potential energy is the total potential energy of a system of charges in an electric field. Electric potential is a scalar quantity, while electric potential energy is a vector quantity.

Can electric potential be negative?

Yes, electric potential can be negative. This means that the electric potential energy is decreasing as the charge moves closer to the source of the electric field. It is also possible for electric potential to be zero, which means that the electric potential energy is constant at all points in the electric field.

Similar threads

Back
Top